ngian
Well-Known Member
Hello everyone
On my latest soap that I made I wanted to use new methods in order to make it and that involves making all the calculations with a pencil and a paper, making a dual alkali soap, and using a full water replacement with vinegar. A while ago topofmurrayhill has posted the "manual" way of calculating ingredients for a recipe without the need of an electronic soap recipe calculator and also how to calculate the excess lye needed for the vinegar in a recipe. Also DeeAnna has shown us how to correct the amount of alkalies really needed in order to make an actual lye discount, eliminating the hidden one found inside the alkali.
I would like to write all these steps once more in one place along with dual alkali calculations that I did on my latest recipe. It is really easy to do them by hand and I feel that someone gains even more control in the entire soaping process.
The only thing we need to know is the SAP value of each oil that our recipe will have. My recipe had
75% Olive Oil (with very low acidity) (600gr)
20% Palm oil and (160gr)
5% Castor oil (40gr)
This recipe could also be named as the no coconut oil soap as I was trying to have bubbles from the addition of the 10% KOH alkali.
So let's start:
1) Calculation of the saponification
The first step to do is to find the quantities of the alkalies needed to saponify our oils based on their SAP values for both NaOH and KOH:
Olive oil: 600 g (SAP - NaOH: 0.135 | KOH: 0.19)
Grams alkali needed for its saponification:
600 x 0,135 = 81gr NaOH
600 x 0,19 = 114gr KOH
Palm oil: 160 g (SAP - NaOH: 0.142 | KOH: 0,199)
Grams alkali needed for its saponification:
160 x 0,142 = 22,72gr NaOH
160 x 0,199 = 31,84gr KOH
Castor oil: 40 g (SAP - NaOH: 0.128 | KOH: 0.18)
Grams alkali needed for its saponification:
40 x 0,128 = 5,12gr NaOH
40 x 0,18 = 7,2gr KOH
So if we sum all grams of soda (NaOH) and all the grams of potash (KOH) will have the total weight of the respective alkalies needed to saponify the oils:
Saponification of all oils only with NaOH: 108,84gr
Saponification of all oil only with KOH: 153,04gr
These weights however are for 100% purity, which is not so in reality. According to the Certification of Analysis documents that the supplier gave me, the NaOH I have in my hands is 98% pure and the KOH is 85% pure, if I store them airtight after opening them and I no further decrease their purity with their exposure to the atmosphere. So the correction of purity is the next step:
2) Correction of the alkali weight needed based on their actual purity
Corrected alkali weight = amount of alkali we calculated above x (100% purity / purity of the alkali that we have in our hands)
NaOH = 108,84 x (100/98) => 108,84 x 1,02 => 111gr
KOH = 153,04 x (100/85) => 153,04 x 1,17 => 179gr
So these weights are what we need to make a soap with an almost real 0% lye discount (I'm writing "almost" as I don't know the real SAP number of the oils I have in hand).
3) Ratio of each alkali in the recipe
The next step is to calculate the weights of each alkali we will use depending on their percentage use in the recipe. In this I decided to use 90% NaOH and 10% KOH, so we will do the following multiplications based on the latest grams of alkalies we found above:
Using NaOH 90%: 111gr x 0,90 = 100g
(If we wanted to use it 95% NaOH in the recipe, we would multiply by 0.95).
Using 10% KOH: 179gr x 0,10 = 17,9gr
(For example if we wanted to use 5% KOH in the recipe, we would multiply by 0.05).
4) Lye discount by 3%
The next step is to calculate the weights of alkali we found above with 3% discount (or as much discount you usually do in your soaps) in both NaOH & KOH:
NaOH discount of 3%, 100g x 0,97 = 97gr
KOH discount of 3%: 17,9gr x 0,97 = 17,3gr
(For example if we wanted to make NaOH / KOH discount of 6% in the recipe then you multiply by 0.94)
5) Calculation of water
After calculating all the alkalies needed for our recipe then we can calculate the water needed based on them. The lye concentration that I will use will be 1,7: 1 (37% lye concentration) which practically means that the amount of water is the weight of the alkali multiplied by 1,7. So according to the latest amounts we got at step 4 we have:
Water needed for NaOH: 97 x 1,7 = 165gr
Water needed for KOH: 17,3 x 1,7 = 30gr
So we will need a total of 165 + 30 = 195gr distilled water to dissolve the alkalies.
But as I said initially, I will use vinegar instead of water in the recipe, to take advantage of the acetic acid that will offer hardness similar to salt. And thus we should still make one more correction in the amount of the alkalies needed so as for them to react with acetic acid without raising further the recipe's lye discount:
6) Alkalies correction due to vinegar
So if you still read me without changing the page in order to read something less boring, we're going to see how to calculate NaOH & KOH we will need to keep the original alkali 3% discount to the recipe.
The vinegar that we have in Greek market is 6% acidity (it is written on its label) and it means that 6% of its weight is acetic acid. The alkalies that are needed to fully react with 1gr of acetic acid is 0,66gr NaOH and 0,92gr KOH respectively. So for the "water" we calculated above we must do the following:
Acetic acid present in vinegar for NaOH: 165gr x 0,06 (6%) = 10g
Additional NaOH needed for the acetic acid: 10 x 0,66 = 6.6 g NaOH
Acetic acid present in vinegar for KOH: 30g x 0,06 (6%) = 1.8 g
Additional KOH needed for the acetic acid: 1,8 x 0,92 = 1.6 g KOH
Corrected NaOH because of vinegar: 97 + 6.6 = 103,6gr
Corrected KOH because of vinegar: 17.3 + 1.6 = 18,9gr
Thus, after all of the six steps above we have:
Olive oil: 600gr
Palm oil: 160gr
Castor oil: 40gr
NaOH: 103,6gr in 165gr of vinegar
KOH: 18,9gr in 30gr of vinegar
The dissolution of the two alkalies can be done in the same vessel with 195gr of vinegar for simplicity. You can follow any steps that you might need for your recipe and any of them that you feel comfortable with (eg. step 2)
If you managed to read me until here, congratulations!
I also used 3% of French Green Clay and Masticha/Sandalwood EOs. In my soap I forgot to calculate the 6) step and as a result I made the soap with almost 10% alkalies discount. The soap seems perfectly hard (I think that acetic acid makes a "better" hardness compared to sodium chloride, along with 37% lye concentration), it is a bit oily in its surface, and as I was cutting it 9 hours after I mold it and CPOP it for 1 hour, it was really hard to cut.
My next project will be to make it once more but with the calculation of the 6) step along with changing the ratio to 85% NaOH and 15% KOH. Vinegar and water discount is something that I like lately in soaps.
On my latest soap that I made I wanted to use new methods in order to make it and that involves making all the calculations with a pencil and a paper, making a dual alkali soap, and using a full water replacement with vinegar. A while ago topofmurrayhill has posted the "manual" way of calculating ingredients for a recipe without the need of an electronic soap recipe calculator and also how to calculate the excess lye needed for the vinegar in a recipe. Also DeeAnna has shown us how to correct the amount of alkalies really needed in order to make an actual lye discount, eliminating the hidden one found inside the alkali.
I would like to write all these steps once more in one place along with dual alkali calculations that I did on my latest recipe. It is really easy to do them by hand and I feel that someone gains even more control in the entire soaping process.
The only thing we need to know is the SAP value of each oil that our recipe will have. My recipe had
75% Olive Oil (with very low acidity) (600gr)
20% Palm oil and (160gr)
5% Castor oil (40gr)
This recipe could also be named as the no coconut oil soap as I was trying to have bubbles from the addition of the 10% KOH alkali.
So let's start:
1) Calculation of the saponification
The first step to do is to find the quantities of the alkalies needed to saponify our oils based on their SAP values for both NaOH and KOH:
Olive oil: 600 g (SAP - NaOH: 0.135 | KOH: 0.19)
Grams alkali needed for its saponification:
600 x 0,135 = 81gr NaOH
600 x 0,19 = 114gr KOH
Palm oil: 160 g (SAP - NaOH: 0.142 | KOH: 0,199)
Grams alkali needed for its saponification:
160 x 0,142 = 22,72gr NaOH
160 x 0,199 = 31,84gr KOH
Castor oil: 40 g (SAP - NaOH: 0.128 | KOH: 0.18)
Grams alkali needed for its saponification:
40 x 0,128 = 5,12gr NaOH
40 x 0,18 = 7,2gr KOH
So if we sum all grams of soda (NaOH) and all the grams of potash (KOH) will have the total weight of the respective alkalies needed to saponify the oils:
Saponification of all oils only with NaOH: 108,84gr
Saponification of all oil only with KOH: 153,04gr
These weights however are for 100% purity, which is not so in reality. According to the Certification of Analysis documents that the supplier gave me, the NaOH I have in my hands is 98% pure and the KOH is 85% pure, if I store them airtight after opening them and I no further decrease their purity with their exposure to the atmosphere. So the correction of purity is the next step:
2) Correction of the alkali weight needed based on their actual purity
Corrected alkali weight = amount of alkali we calculated above x (100% purity / purity of the alkali that we have in our hands)
NaOH = 108,84 x (100/98) => 108,84 x 1,02 => 111gr
KOH = 153,04 x (100/85) => 153,04 x 1,17 => 179gr
So these weights are what we need to make a soap with an almost real 0% lye discount (I'm writing "almost" as I don't know the real SAP number of the oils I have in hand).
3) Ratio of each alkali in the recipe
The next step is to calculate the weights of each alkali we will use depending on their percentage use in the recipe. In this I decided to use 90% NaOH and 10% KOH, so we will do the following multiplications based on the latest grams of alkalies we found above:
Using NaOH 90%: 111gr x 0,90 = 100g
(If we wanted to use it 95% NaOH in the recipe, we would multiply by 0.95).
Using 10% KOH: 179gr x 0,10 = 17,9gr
(For example if we wanted to use 5% KOH in the recipe, we would multiply by 0.05).
4) Lye discount by 3%
The next step is to calculate the weights of alkali we found above with 3% discount (or as much discount you usually do in your soaps) in both NaOH & KOH:
NaOH discount of 3%, 100g x 0,97 = 97gr
KOH discount of 3%: 17,9gr x 0,97 = 17,3gr
(For example if we wanted to make NaOH / KOH discount of 6% in the recipe then you multiply by 0.94)
5) Calculation of water
After calculating all the alkalies needed for our recipe then we can calculate the water needed based on them. The lye concentration that I will use will be 1,7: 1 (37% lye concentration) which practically means that the amount of water is the weight of the alkali multiplied by 1,7. So according to the latest amounts we got at step 4 we have:
Water needed for NaOH: 97 x 1,7 = 165gr
Water needed for KOH: 17,3 x 1,7 = 30gr
So we will need a total of 165 + 30 = 195gr distilled water to dissolve the alkalies.
But as I said initially, I will use vinegar instead of water in the recipe, to take advantage of the acetic acid that will offer hardness similar to salt. And thus we should still make one more correction in the amount of the alkalies needed so as for them to react with acetic acid without raising further the recipe's lye discount:
6) Alkalies correction due to vinegar
So if you still read me without changing the page in order to read something less boring, we're going to see how to calculate NaOH & KOH we will need to keep the original alkali 3% discount to the recipe.
The vinegar that we have in Greek market is 6% acidity (it is written on its label) and it means that 6% of its weight is acetic acid. The alkalies that are needed to fully react with 1gr of acetic acid is 0,66gr NaOH and 0,92gr KOH respectively. So for the "water" we calculated above we must do the following:
Acetic acid present in vinegar for NaOH: 165gr x 0,06 (6%) = 10g
Additional NaOH needed for the acetic acid: 10 x 0,66 = 6.6 g NaOH
Acetic acid present in vinegar for KOH: 30g x 0,06 (6%) = 1.8 g
Additional KOH needed for the acetic acid: 1,8 x 0,92 = 1.6 g KOH
Corrected NaOH because of vinegar: 97 + 6.6 = 103,6gr
Corrected KOH because of vinegar: 17.3 + 1.6 = 18,9gr
Thus, after all of the six steps above we have:
Olive oil: 600gr
Palm oil: 160gr
Castor oil: 40gr
NaOH: 103,6gr in 165gr of vinegar
KOH: 18,9gr in 30gr of vinegar
The dissolution of the two alkalies can be done in the same vessel with 195gr of vinegar for simplicity. You can follow any steps that you might need for your recipe and any of them that you feel comfortable with (eg. step 2)
If you managed to read me until here, congratulations!
I also used 3% of French Green Clay and Masticha/Sandalwood EOs. In my soap I forgot to calculate the 6) step and as a result I made the soap with almost 10% alkalies discount. The soap seems perfectly hard (I think that acetic acid makes a "better" hardness compared to sodium chloride, along with 37% lye concentration), it is a bit oily in its surface, and as I was cutting it 9 hours after I mold it and CPOP it for 1 hour, it was really hard to cut.
My next project will be to make it once more but with the calculation of the 6) step along with changing the ratio to 85% NaOH and 15% KOH. Vinegar and water discount is something that I like lately in soaps.
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